Prove that the arithmetic sequence with first term 1/3 and common difference 1/6 contains all natural numbers.
For the sequence,
a = ![]()
d = ![]()
an = a + (n-1) × d
Let p be any natural number
p =
+ (n-1) × ![]()
=
+ ![]()
= ![]()
6p = n + 1
Thus, for every natural number p, a value of n exists.
Hence, this arithmetic sequence contains all natural numbers.
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